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Mth 277
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Curve of Intersection
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Find parametric equations for the curve of the intersection of the elliptical cylinder x^2 / 16 + z^2 / 4 =1 and the parabolic cylinder z = .5 x^2
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The question that I have for this is I am trying to solve it and I am confused as to how to find the parametric equations without a point given and doing the cross product of that. Also, since there are only 2 variables I do not think we do the cross product. My best bet is to graph them and then find a point that they intersect at. In that case, we get (-1.88,1.88). Do we use this as the point for the question and use 0 for the y in the cross product? Then, we use this for the parametric equations?
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A graphing calculator can give you an indication, but not a solution.
If you treat the two equations as simultaneous equations you get a solution.
I believe I also emailed you a document with a solution to this problem, and hints\solutions on others. The method used in the solution to this problem generalizes to problems of this nature that couldn't in any case be easily solved with the use of a calculator.
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#$&*
Mth 277
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Unit Normal Vector
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Find the velocity, the unit tangent vector, the acceleration and the unit normal vectors for the position function r(t) = (t * cos t) i - sqrt(t) j - t * sin(2t) k.
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I understand that the velocity is the derivative of the position and the acceleration is the derivative of the velocity. I got those derivatives taken care of no problem. I also understand that the unit tangent vector is the velocity vector divided by the absolute value of the velocity vector. I was trying to figure out what the equation meant by the absolute value. Do we just switch each sign of the velocity vector components that are negative to positive for the i j and k? And wouldn't this always be equal to 1 or -1 since the top is being divided by the same thing on the bottom but signs are just switching?
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The magnitude of the velocity vector for this position function is not 1. You find the magnitude by using the Pythagorean Theorem. The squared components will of course all be positive.
With respect to terminology, I understand what you mean when you say 'absoolute value', but that term isn't completely correct when applied to vectors. The correct term is 'magnitude', and using that term can eliminate some confusion.
Dividing a vector with three components by its magnitude gives you another vector with three components. If you divide a vector by a scalar you get a vector; you can't get 1 or -1, which are scalars.
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Also, I was wondering about the unit normal vector. I know what the normal vector is and the unit vector. But I was wondering what we use to find the unit normal vector? I have read in the book and been unable to find examples.
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To get the unit normal you divide the normal vector by its magnitude.
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